[python] How to normalize a 2-dimensional numpy array in python less verbose?

Given a 3 times 3 numpy array

a = numpy.arange(0,27,3).reshape(3,3)

# array([[ 0,  3,  6],
#        [ 9, 12, 15],
#        [18, 21, 24]])

To normalize the rows of the 2-dimensional array I thought of

row_sums = a.sum(axis=1) # array([ 9, 36, 63])
new_matrix = numpy.zeros((3,3))
for i, (row, row_sum) in enumerate(zip(a, row_sums)):
    new_matrix[i,:] = row / row_sum

There must be a better way, isn't there?

Perhaps to clearify: By normalizing I mean, the sum of the entrys per row must be one. But I think that will be clear to most people.

In case you are trying to normalize each row such that its magnitude is one (i.e. a row's unit length is one or the sum of the square of each element in a row is one):

import numpy as np

a = np.arange(0,27,3).reshape(3,3)

result = a / np.linalg.norm(a, axis=-1)[:, np.newaxis]
# array([[ 0.        ,  0.4472136 ,  0.89442719],
#        [ 0.42426407,  0.56568542,  0.70710678],
#        [ 0.49153915,  0.57346234,  0.65538554]])

Verifying:

np.sum( result**2, axis=-1 )
# array([ 1.,  1.,  1.]) 

You could use built-in numpy function: np.linalg.norm(a, axis = 1, keepdims = True)

Here is one more possible way using reshape:

a_norm = (a/a.sum(axis=1).reshape(-1,1)).round(3)
print(a_norm)

Or using None works too:

a_norm = (a/a.sum(axis=1)[:,None]).round(3)
print(a_norm)

Output:

array([[0.   , 0.333, 0.667],
       [0.25 , 0.333, 0.417],
       [0.286, 0.333, 0.381]])

normed_matrix = normalize(input_data, axis=1, norm='l1')
print(normed_matrix)

where input_data is the name of your 2D array

Scikit-learn offers a function normalize() that lets you apply various normalizations. The "make it sum to 1" is called L1-norm. Therefore:

from sklearn.preprocessing import normalize

matrix = numpy.arange(0,27,3).reshape(3,3).astype(numpy.float64)
# array([[  0.,   3.,   6.],
#        [  9.,  12.,  15.],
#        [ 18.,  21.,  24.]])

normed_matrix = normalize(matrix, axis=1, norm='l1')
# [[ 0.          0.33333333  0.66666667]
#  [ 0.25        0.33333333  0.41666667]
#  [ 0.28571429  0.33333333  0.38095238]]

Now your rows will sum to 1.

it appears that this also works

def normalizeRows(M):
    row_sums = M.sum(axis=1)
    return M / row_sums

Or using lambda function, like

>>> vec = np.arange(0,27,3).reshape(3,3)
>>> import numpy as np
>>> norm_vec = map(lambda row: row/np.linalg.norm(row), vec)

each vector of vec will have a unit norm.

You could also use matrix transposition:

(a.T / row_sums).T

I think you can normalize the row elements sum to 1 by this: new_matrix = a / a.sum(axis=1, keepdims=1). And the column normalization can be done with new_matrix = a / a.sum(axis=0, keepdims=1). Hope this can hep.

I think this should work,

a = numpy.arange(0,27.,3).reshape(3,3)

a /=  a.sum(axis=1)[:,numpy.newaxis]